19 replies to this topic

[Prime]

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

 

 

Here's a winning strategy

Spoiler for

Pick to be the first player and choose 5 as the first number. Let the second player's first choice be S.

If S is 1, 9, 2, 6, or 7, then player 1's 2nd, 3rd, and 4th moves should be among the following set (3,4,8)
If S is 3, 4, or 8, then player 1's 2nd, 3rd, and 4th moves should be among the following set (2, 6, 7). The key is that after playing S, player 2 would be constrained to block player 1's move from getting 15.

Example. Let say player 1 choose 5, and player 2 chooses 2. The game would go as follows

Player 1- 1st choice: 5
Player 2- 1st choice: 2
Player 1- 2nd choice: 3
Player 2- 2nd choice: 7 (have to choose 7 to prevent a player 1 win)
Player 1- 3rd choice: 4
Player 2- 3rd choice: 6 (have to choose 6 to prevent a player 1 win)
Player 1- 4th choice: 8 (WIN- 3 + 4 + 8 = 15 )

In this example, doesn't 2nd player win on his 3rd move having (2,7,6)?

Edited by Prime, 29 January 2013 - 06:27 PM.

[bushindo]

 

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

 

 

Here's a winning strategy

Spoiler for

Pick to be the first player and choose 5 as the first number. Let the second player's first choice be S.

If S is 1, 9, 2, 6, or 7, then player 1's 2nd, 3rd, and 4th moves should be among the following set (3,4,8)
If S is 3, 4, or 8, then player 1's 2nd, 3rd, and 4th moves should be among the following set (2, 6, 7). The key is that after playing S, player 2 would be constrained to block player 1's move from getting 15.

Example. Let say player 1 choose 5, and player 2 chooses 2. The game would go as follows

Player 1- 1st choice: 5
Player 2- 1st choice: 2
Player 1- 2nd choice: 3
Player 2- 2nd choice: 7 (have to choose 7 to prevent a player 1 win)
Player 1- 3rd choice: 4
Player 2- 3rd choice: 6 (have to choose 6 to prevent a player 1 win)
Player 1- 4th choice: 8 (WIN- 3 + 4 + 8 = 15 )

In this example, doesn't 2nd player win on his 3rd move having (2,7,6)?

 

Shucks, back to the drawing board.

[CaptainEd]

Yes, now you guys have made it pretty clear

Spoiler for Funny 3-D tic tac toe

If Player A plays 5 and B plays 6, I think B has time to respond to A's forces, or has time to force A to respond, resulting in a draw.

 

There are 8 3-chip combinations that sum to 15: 168, 159, 267, 258, 249, 357, 348, 456.

Of these, the most useful chip is 5, as it has 4 combinations, while the other chips only pertain to 2 or 3.

So A chooses 5. This knocks out those 4 combinations from B's possibilities.

 

If B chooses 6, this knocks out 3 combinations from A's possibilities.

If A chooses 4, this knocks out all but two of B's possibilities. Even so, as far as I can tell, B can prevent A from creating a fork.

 

I can't give an exhaustive list, or a principle/algorithm that is easy to follow.

[Prime]

Spoiler for Not an answer

All possible game variants may be traversed by a computer program. I suspect the result would be a draw. A smarter computer program would utilize a "forced move" concept. A simpler program could be the old mini-max algorithm.

[bonanova]

Yes, now you guys have made it pretty clear

Spoiler for Funny 3-D tic tac toe

If Player A plays 5 and B plays 6, I think B has time to respond to A's forces, or has time to force A to respond, resulting in a draw.
 
There are 8 3-chip combinations that sum to 15: 168, 159, 267, 258, 249, 357, 348, 456.
Of these, the most useful chip is 5, as it has 4 combinations, while the other chips only pertain to 2 or 3.
So A chooses 5. This knocks out those 4 combinations from B's possibilities.
 
If B chooses 6, this knocks out 3 combinations from A's possibilities.
If A chooses 4, this knocks out all but two of B's possibilities. Even so, as far as I can tell, B can prevent A from creating a fork.
 
I can't give an exhaustive list, or a principle/algorithm that is easy to follow.

This is correct.

Can you draw a picture?

[CaptainEd]

Spoiler for I would like to draw a good picture

I'm not happy with my pictures, they do not capture my tictactoe feeling
The only 3-chip combinations that add to 15:

 
The A and B below indicate that, at this stage of the game, both A and B have this possibility available to them. As a player chooses one of the chips, the other player loses that possibility.

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
A A A A A A A A
B B B B B B B B

 
Note that there are four combinations with 5, three combinations with each of 2,4,6,8, and two combinations with each of 1,3,7,9

 

(1) 5, A chooses 5, because it has the most combinations

 

1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
A A A A A A A A
B   B   B   B 

 
(2) B chooses 6, (it knocks out 3 of A's combinations--I believe 4 would have been equivalent)

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A   A A A A 
B   B   B   B

 
(3) A chooses 4, because it knocks out half of B's possibilities, while advancing two of A's own possibilities

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A   A A A A 
B   B      

 
At this point, A has 54 and B has 6.
B had better not choose 7, appealing though it be, as A is forced to choose 2, which creates a fork, with A threatening both 258 and 249. I believe the choice of 2 has the similar failure.
B's better choice is 8.

 
(4) B chooses 8, threatening 168

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A     A A   
B   B      

 
(5) A has no immediate win, so is forced to choose 1, threatening 159

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A     A A   
    B      

 
A has 145, B has 68

 
(6) B is forced to choose 9

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
          A   
    B      

 
A has 145, B has 689

 
(7) A chooses 7, threatening 357 and blocking B's last combination (267)
(8) B chooses 3, blocking A's last combination (357)

 
-----
After the first three moves (A5, B6, A4), I think it is clear that both players can force a draw (or can mess up and lose)

 
One could question whether one of the three first moves could be replaced by more powerful ones. But each move seems to have a good rationale...

Edited by bonanova, 30 January 2013 - 05:16 PM.

[bonanova]

Great. You win.

But what I really meant was can you draw a magic 3x3 picture?

[CaptainEd]

Great. You win.

But what I really meant was can you draw a magic 3x3 picture?

Oh, woops! Uh, no...especially on a day when I can't even code a spoiler

[bonanova]

8 1 6
3 5 7
4 9 2

Does that help?

[CaptainEd]

 Rather easy when you put it that way...